From bfdbc1d5718e11e3f77a04623a1a9593c7ce5da0 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Tue, 12 Jul 2022 11:08:29 +0200
Subject: more writing.

---
 frsb_kac-rice.tex | 87 ++++++++++++++++++++++++++++++-------------------------
 1 file changed, 47 insertions(+), 40 deletions(-)

(limited to 'frsb_kac-rice.tex')

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index 05eccb3..e42046d 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -368,7 +368,7 @@ Therefore in that limit its spectrum is given by the Wigner semicircle with radi
 \end{equation}
 The spectrum of the Hessian $\operatorname{Hess}H(\mathbf s,\mu)$ is the same
 semicircle shifted by $\mu$, or $\rho(\lambda+\mu)$.  The stability parameter
-$\mu$ thus fixes the center of the spectrum of the Hessian. The value
+$\mu$ thus fixes the center of the spectrum of the Hessian. The semicircle radius
 $\mu_m=\sqrt{4f''(1)}$ is a kind of threshold. When $\mu$ is taken to be within
 the range $\pm\mu_m$, the critical points have index density
 \begin{equation}
@@ -1018,7 +1018,12 @@ Though $E_\mathrm{alg}$ looks quite close to the energy at which dominant
 saddles transition from 1RSB to RS, they differ by roughly $10^{-3}$, as
 evidenced by the numbers cited above. Likewise, though $\langle E\rangle_1$
 looks very close to $E_\mathrm{max}$, where the 1RSB transition line
-terminates, they too differ.
+terminates, they too differ. The fact that $E_\mathrm{alg}$ is very slightly
+below the place where most saddle transition to 1RSB is suggestive; we
+speculate that an analysis of the typical minima connected to these saddles by
+downward trajectories will coincide with the algorithmic limit. An analysis of
+the typical nearby minima or the typical downward trajectories from these
+saddles at 1RSB is warranted \cite{Ros_2019_Complex, Ros_2021_Dynamical}.
 
 \begin{figure}
   \centering
@@ -1167,14 +1172,13 @@ goes to zero with finite slopes.
   } \label{fig:24.func}
 \end{figure}
 
-
 \section{Interpretation}
 \label{sec:interpretation}
 
 Let $\langle A\rangle$ be the average of $A$ over stationary points with given $E$ and $\mu^*$, i.e.,
 \begin{equation}
   \langle A\rangle
-  =\frac1{\mathcal N}\sum_{\sigma}A(\mathbf s_\sigma)
+  =\frac1{\mathcal N}\sum_{\mathbf s\in\mathcal S}A(\mathbf s)
   =\frac1{\mathcal N}
   \int d\nu(\mathbf s)\,A(\mathbf s)
 \end{equation}
@@ -1183,40 +1187,43 @@ with
   d\nu(\mathbf s)=d\mathbf s\,d\mu\,\delta\big(\tfrac12(\|\mathbf s\|^2-N)\big)\,\delta\big(\nabla H(\mathbf s,\mu)\big)\,\big|\det\operatorname{Hess}H(\mathbf s,\mu)\big|
     \delta\big(NE-H(\mathbf s)\big)\delta\big(N\mu^*-\operatorname{Tr}\operatorname{Hess}H(\mathbf s,\mu)\big)
 \end{equation}
-the Kac--Rice measure. Note that this definition of the angle brackets is not
-the same as that used in \S\ref{subsec:expansion} The fields $C$, $R$, and $D$
-defined in \eqref{eq:fields} can be related to certain averages of this type.
+the Kac--Rice measure. Note that this definition of the angle brackets, which
+is in analogy with the typical equilibrium average, is not the same as that
+used in \S\ref{subsec:expansion} for averaging over the off-diagonal elements
+of a hierarchical matrix. The fields $C$, $R$, and $D$ defined in
+\eqref{eq:fields} can be related to certain averages of this type.
 
 \subsection{\textit{C}: distribution of overlaps}
 
 First,
 consider $C$, which has an interpretation nearly identical to that of Parisi's
-$Q$ matrix of overlaps. It can be shown that its off-diagonal corresponds to
-the probability distribution of the overlaps between stationary points $P(q)$.
-Let $\mathcal S$ be the set of all stationary points with given energy density
-and index. Then
+$Q$ matrix of overlaps in the equilibrium case. Its off-diagonal corresponds to
+the probability distribution $P(q)$ of the overlaps $q=(\mathbf s_1\cdot\mathbf
+s_2)/N$ between stationary points.  Let $\mathcal S$ be the set of all
+stationary points with given energy density and index. Then
 \begin{equation}
-P(q)=\frac1{\mathcal N^2}\sum_{\mathbf s_1\in\mathcal S}\sum_{\mathbf s_2\in\mathcal S}\delta\left(\frac{\mathbf s_1\cdot\mathbf s_2}N-q\right)
+P(q)\equiv\frac1{\mathcal N^2}\sum_{\mathbf s_1\in\mathcal S}\sum_{\mathbf s_2\in\mathcal S}\delta\left(\frac{\mathbf s_1\cdot\mathbf s_2}N-q\right)
 \end{equation}
-{\em This is the probability that two stationary points randomly drawn from the ensemble
-of stationary points happen to be at overlap $q$}
-
-%It is straightforward to show that moments of this distribution are related to
-%certain averages of the form. 
-These are evaluated for a given energy, index, etc, but
-we shall omit these subindices for simplicity.
+{\em This is the probability that two stationary points uniformly drawn from the ensemble
+of all stationary points with fixed $E$ and $\mu^*$ happen to be at overlap $q$.}
+Though these are evaluated for a given energy, index, etc, we shall omit these
+subindices for simplicity.
 
+The moments of this distribution $q^{(p)}$ are given by
 \begin{equation}
   \begin{aligned}
     q^{(p)}
-    &\equiv \frac1{N^p}\sum_{i_1\cdots i_p}\langle s_{i_1}\cdots s_{i_p}\rangle\langle s_{i_1}\cdots s_{i_p}\rangle
+    &\equiv\int_0^1dq\,q^pP(q)
+    =\frac1{N^p}\sum_{i_1\cdots i_p}\langle s_{i_1}\cdots s_{i_p}\rangle\langle s_{i_1}\cdots s_{i_p}\rangle
     =\frac1{N^p} \; \frac{1}{{\cal{N}}^2} \left\{ \sum_{{\mathbf s}_1,{\mathbf s}_2}\; \sum_{i_1\cdots i_p} s^1_{i_1}\cdots s^1_{i_p} s^2_{i_1}\cdots s^2_{i_p}\right\} \\
-    &=\frac1{N^p} \; \frac{1}{{\cal{N}}^2} \left\{\sum_{{\mathbf s}_1,{\mathbf s}_2} (\mathbf s_1\cdot\mathbf s_2)^p\right\}
-    =\frac1{N^p} \lim_{n\to0} \left\{\sum_{{\mathbf s}_1,{\mathbf s}_2,\ldots,\mathbf s_n} (\mathbf s_1\cdot\mathbf s_2)^p\right\}
+    &=\frac{1}{{\mathcal{N}}^2} \left\{\sum_{{\mathbf s}_1,{\mathbf s}_2}\left(\frac{\mathbf s_1\cdot\mathbf s_2}N\right)^p\right\}
+    =\lim_{n\to0} \left\{\sum_{{\mathbf s}_1,{\mathbf s}_2,\ldots,\mathbf s_n} \left(\frac{\mathbf s_1\cdot\mathbf s_2}N\right)^p\right\}
   \end{aligned}
 \end{equation}
-The $(n-2)$ extra replicas providing the normalization.
-Replacing the sums over stationary points with integrals over the Kac--Rice measure, the average over disorder, and again, for given energy and index, gives
+The $(n-2)$ extra replicas provide the normalization, with
+$\lim_{n\to0}\mathcal N^{n-2}=\mathcal N^{-2}$.  Replacing the sums over
+stationary points with integrals over the Kac--Rice measure, the average over
+disorder (again, for fixed energy and index) gives
 \begin{equation}
   \begin{aligned}
     \overline{q^{(p)}} &=\overline{\frac1{N^p}\sum_{i_1\cdots i_p}\langle s_{i_1}\cdots s_{i_p}\rangle\langle s_{i_1}\cdots s_{i_p}\rangle}
@@ -1229,7 +1236,7 @@ Replacing the sums over stationary points with integrals over the Kac--Rice meas
 In the last line, we have used that there is nothing special about replicas one and two.
 Using the Parisi ansatz, evaluating by saddle point  {\em summing over all the $n(n-1)$ saddles related by permutation} we then have
 \begin{equation}
- \overline{q^{(p)}}=\int_0^1 dx\,c^p(x) = \int dq\,q^p P(q) \qquad ; \qquad {\mbox{with}} \qquad P(q)=\frac{dx}{dq}
+  \overline{q^{(p)}}=\int_0^1 dx\,c^p(x) = \int_0^1 dq\,q^p P(q) \qquad \qquad {\mbox{concluding}} \qquad P(q)=\frac{dx}{dq}=\left(\frac{dc}{dx}\right)^{-1}\bigg|_{c(x)=q}
 \end{equation}
 The appeal of Parisi to properties of pure states is unnecessary here, since
 the stationary points are points.
@@ -1238,10 +1245,10 @@ With this
 established, we now address what it means for $C$ to have a nontrivial
 replica-symmetry broken structure. When $C$ is replica symmetric, drawing two
 stationary points at random will always lead to the same overlap. In the case
-when there is no linear field, they will always have overlap zero, because the
-second point will almost certainly lie on the equator of the sphere with
-respect to the first. Though other stationary points exist nearby the first
-one, they are exponentially fewer and so will be picked with vanishing
+when there is no linear field and $q_0=0$, they will always have overlap zero,
+because the second point will almost certainly lie on the equator of the sphere
+with respect to the first. Though other stationary points exist nearby the
+first one, they are exponentially fewer and so will be picked with vanishing
 probability in the thermodynamic limit.
 
 When $C$ is replica-symmetry broken, there is a nonzero probability of picking
@@ -1250,21 +1257,21 @@ imagining the level sets of the Hamiltonian in this scenario. If the level sets
 are disconnected but there are exponentially many of them distributed on the
 sphere, one will still find zero average overlap. However, if the disconnected
 level sets are \emph{few}, i.e., less than order $N$, then it is possible to
-draw two stationary points from the same set. Therefore, the picture in this
-case is of few, large basins each containing exponentially many stationary
-points.
+draw two stationary points from the same set with nonzero probability.
+Therefore, the picture in this case is of few, large basins each containing
+exponentially many stationary points.
 
 \subsection{A concrete example}
 
 One can construct a schematic 2RSB model from two 1RSB models.
-Consider two independent pure $p$ spin models $H_{p_1}({\mathbf s})$ and
-$H_{p_2}({\mathbf \sigma})$ of sizes  $N$, and couple them weakly with
-$\varepsilon \;
-{\mathbf \sigma} \cdot {\mathbf s}$. The landscape of the pure models is much
-simpler than that of the mixed because, in these models, fixing the stability
-$\mu$ is equivalent to fixing the energy: $\mu=pE$. This implies that at each
-energy level there is only one type of stationary point. Therefore, for the
-pure models our formulas for the complexity and its Legendre transforms are
+Consider two independent pure models with $p_1$ and $p_2$ couplings,
+respectively, with energies $H_{p_1}({\mathbf s})$ and $H_{p_2}({\mathbf
+\sigma})$ of sizes  $N$, and couple them weakly with $\varepsilon \; {\mathbf
+\sigma} \cdot {\mathbf s}$. The landscape of the pure models is much simpler
+than that of the mixed because, in these models, fixing the stability $\mu$ is
+equivalent to fixing the energy: $\mu=pE$. This implies that at each energy
+level there is only one type of stationary point. Therefore, for the pure
+models our formulas for the complexity and its Legendre transforms are
 functions of one variable only, $E$, and each instance of $\mu^*$ inside mus be
 replaced with $pE$.
 
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